hdu 4726 Kia's Calculation

Kia's Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 481    Accepted Submission(s): 133

Problem Description

Doctor Ghee is teaching Kia how to calculate the sum of two integers. But Kia is so careless and alway forget to carry a number when the sum of two digits exceeds 9. For example, when she calculates 4567+5789, she will get 9246, and for 1234+9876, she will get 0. Ghee is angry about this, and makes a hard problem for her to solve:
Now Kia has two integers A and B, she can shuffle the digits in each number as she like, but leading zeros are not allowed. That is to say, for A = 11024, she can rearrange the number as 10124, or 41102, or many other, but 02411 is not allowed.
After she shuffles A and B, she will add them together, in her own way. And what will be the maximum possible sum of A "+" B ?

 

Input

The rst line has a number T (T <= 25) , indicating the number of test cases.
For each test case there are two lines. First line has the number A, and the second line has the number B.
Both A and B will have same number of digits, which is no larger than 10⁶, and without leading zeros.

 

Output

For test case X, output "Case #X: " first, then output the maximum possible sum without leading zeros.

 

Sample Input

159583036

 

Sample Output

Case #1: 8984

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

 

Recommend

zhuyuanchen520

错了好多次了,用贪心的方法,从大数字到小数字来找,这样才不会超 时!

#include <stdio.h>#include <string.h>#include <iostream>using namespace std;int p[2][10];char str1[1000050],str2[1000050];int main (){    int tcase,a,b,num1,num2,i,j,k,tt=1;    scanf("%d",&tcase);    gets(str1);    while(tcase--)    {        gets(str1);        gets(str2);        num1=0;        memset(p,0,sizeof(p));        for(num1=0;str1[num1]!='\0';num1++)        {            p[0][str1[num1]-'0']++;            p[1][str2[num1]-'0']++;        }         printf("Case #%d: ",tt++);        bool flag=true,flag2=true;        int li;        for(i=0,li=9;i<num1;i++)        {            flag=true;            for(;li>=0&&flag;li--)            {               for(j=0;j<=9&&flag;j++)               {                  k=(li-j+10)%10;                  if(!i&&(!j||!k))                  {                    continue;                  }                  if(p[0][j]&&p[1][k])                  {                      p[0][j]--;p[1][k]--;                      flag=false;                      if(flag2&&li!=0)                      printf("%d",li),flag2=false;                      else if(!flag2)                      printf("%d",li);                      goto my;                  }               }            }            my:;            if(!i)            li=9;        }        if(flag2)        {            printf("0\n");        }        else        printf("\n");    }    return 0;}